TL;DR
Use:
try:
i = numpy.where(arr == "string1")[0][0]
except IndexError:
# handle the case where "string1" was not found in arr
or
indices = list(numpy.where(arr == "string1")[0])
Details
Finding elements in NumPy arrays is not intuitive the first time you try to do it.
Let's decompose the operation:
>>> arr = numpy.array(["string1","string2","string3"])
>>> arr == "string1"
array([ True, False, False])
Notice how just doing arr == "string1" is already doing the search: it's returning an array of booleans of the same shape as arr telling us where the condition is true.
Then, you're using numpy.where which, when used with only one parameter (the condition), returns where its input is non-zero. With booleans, that means non false.
>>> numpy.where(numpy.array([ True, False, False]))
(array([0], dtype=int64),)
>>> numpy.where(arr == "string1")
(array([0], dtype=int64),)
It's not quite clear to my where this gives you a tuple of arrays for a 1-D input, but when you use this syntax with a 2-d input, it makes more sense.
In any case, what you're getting here is a tuple containing a list of indices where the condition matches. Notice it has to be a list, because you might have multiple matches.
For your code, you want numpy.where(arr == "string1")[0][0], because you know "string1" occurs in the list, but the inner list may also contain zero or more than one values, depending on how many times the string is found.
>>> arr2 = numpy.array(["string1","string2","string3","string1", "string3"])
>>> numpy.where(arr2 == "foo")
(array([], dtype=int64),)
>>> numpy.where(arr2 == "string3")
(array([2, 4], dtype=int64),)
So when you want to use these indices, you should simply treat numpy.where(arr == "string1")[0] as a list (it's really a 1-D array, though) and continue from there.
Now, just using numpy.where(arr == "some string")[0][0] is risky, because it will throw an IndexError exception if the string is not found in arr. If you really want to do that, do it in a try/except block.
If you need the list of indices as a Python list, you can do this:
indices = list(numpy.where(arr == "string1")[0])
TL;DR
Use:
try:
i = numpy.where(arr == "string1")[0][0]
except IndexError:
# handle the case where "string1" was not found in arr
or
indices = list(numpy.where(arr == "string1")[0])
Details
Finding elements in NumPy arrays is not intuitive the first time you try to do it.
Let's decompose the operation:
>>> arr = numpy.array(["string1","string2","string3"])
>>> arr == "string1"
array([ True, False, False])
Notice how just doing arr == "string1" is already doing the search: it's returning an array of booleans of the same shape as arr telling us where the condition is true.
Then, you're using numpy.where which, when used with only one parameter (the condition), returns where its input is non-zero. With booleans, that means non false.
>>> numpy.where(numpy.array([ True, False, False]))
(array([0], dtype=int64),)
>>> numpy.where(arr == "string1")
(array([0], dtype=int64),)
It's not quite clear to my where this gives you a tuple of arrays for a 1-D input, but when you use this syntax with a 2-d input, it makes more sense.
In any case, what you're getting here is a tuple containing a list of indices where the condition matches. Notice it has to be a list, because you might have multiple matches.
For your code, you want numpy.where(arr == "string1")[0][0], because you know "string1" occurs in the list, but the inner list may also contain zero or more than one values, depending on how many times the string is found.
>>> arr2 = numpy.array(["string1","string2","string3","string1", "string3"])
>>> numpy.where(arr2 == "foo")
(array([], dtype=int64),)
>>> numpy.where(arr2 == "string3")
(array([2, 4], dtype=int64),)
So when you want to use these indices, you should simply treat numpy.where(arr == "string1")[0] as a list (it's really a 1-D array, though) and continue from there.
Now, just using numpy.where(arr == "some string")[0][0] is risky, because it will throw an IndexError exception if the string is not found in arr. If you really want to do that, do it in a try/except block.
If you need the list of indices as a Python list, you can do this:
indices = list(numpy.where(arr == "string1")[0])
Method 1 : This will give you all the indices in a list : [0,3]
using np.where(arr == "string1")[0]
and then convert to tuple
import numpy as np
arr = np.array(["string1", "string2", "string3", "string1"])
indices_tuple = tuple(np.where(arr == "string1")[0])
print(indices_tuple)
To get the row numbers where at least one item is larger than 15:
>>> np.where(np.any(e>15, axis=1))
(array([1, 2], dtype=int64),)
You can use the nonzero function. it returns the nonzero indices of the given input.
Easy Way
>>> (e > 15).nonzero()
(array([1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]), array([6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]))
to see the indices more cleaner, use transpose method:
>>> numpy.transpose((e>15).nonzero())
[[1 6]
[1 7]
[1 8]
[1 9]
[2 0]
...
Not Bad Way
>>> numpy.nonzero(e > 15)
(array([1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]), array([6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]))
or the clean way:
>>> numpy.transpose(numpy.nonzero(e > 15))
[[1 6]
[1 7]
[1 8]
[1 9]
[2 0]
...